Question

in line integral how can we change the vector field(x,y) into parametrized(x(t),y(t)) form, since the parametrization is only given for the curve as x=f(t);y=f(t) and not the vector field

Answer 1

I think you just did, replace x and y with f1(t) and f2(t)

Answer 2

well you are given F(x,y)=x+y (x,y)=(t+1,t)
\[\int\limits_{\zeta C}^{} F*r'(t)dt\]
then r(t)=(t+1,t) the r'(t)=(1,) ds=||r'(t)||=sqrt(1+1)
\[\int\limits_{\zeta C}^{} t+1+t \sqrt{2}dt\]
because x=t+1
y=t you sub those in
and if 0=<t=<1 then you just evaluate it

Answer 3

if your given two points (a,b,c)=P and (x,y,z)=Q
PQ(vector sign on top)=(x-a)it+(y-b)jt+(z-c)kt
x=a+(x-a)t y=b+(y-b)t+c(c-z)t and your t should always go from [0,1]